YES

Problem:
 f(f(x)) -> f(c(f(x)))
 f(f(x)) -> f(d(f(x)))
 g(c(x)) -> x
 g(d(x)) -> x
 g(c(h(0()))) -> g(d(1()))
 g(c(1())) -> g(d(h(0())))
 g(h(x)) -> g(x)

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> f#(c(f(x)))
   f#(f(x)) -> f#(d(f(x)))
   g#(c(h(0()))) -> g#(d(1()))
   g#(c(1())) -> g#(d(h(0())))
   g#(h(x)) -> g#(x)
  TRS:
   f(f(x)) -> f(c(f(x)))
   f(f(x)) -> f(d(f(x)))
   g(c(x)) -> x
   g(d(x)) -> x
   g(c(h(0()))) -> g(d(1()))
   g(c(1())) -> g(d(h(0())))
   g(h(x)) -> g(x)
  Usable Rule Processor:
   DPs:
    f#(f(x)) -> f#(c(f(x)))
    f#(f(x)) -> f#(d(f(x)))
    g#(c(h(0()))) -> g#(d(1()))
    g#(c(1())) -> g#(d(h(0())))
    g#(h(x)) -> g#(x)
   TRS:
    f(f(x)) -> f(c(f(x)))
    f(f(x)) -> f(d(f(x)))
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [g#](x0) = x0 + 0,
     
     [f#](x0) = x0 + 2,
     
     [1] = 0,
     
     [h](x0) = 1x0 + 2,
     
     [0] = 0,
     
     [d](x0) = 0,
     
     [c](x0) = 1,
     
     [f](x0) = x0 + 3
    orientation:
     f#(f(x)) = x + 3 >= 2 = f#(c(f(x)))
     
     f#(f(x)) = x + 3 >= 2 = f#(d(f(x)))
     
     g#(c(h(0()))) = 1 >= 0 = g#(d(1()))
     
     g#(c(1())) = 1 >= 0 = g#(d(h(0())))
     
     g#(h(x)) = 1x + 2 >= x + 0 = g#(x)
     
     f(f(x)) = x + 3 >= 3 = f(c(f(x)))
     
     f(f(x)) = x + 3 >= 3 = f(d(f(x)))
    problem:
     DPs:
      
     TRS:
      f(f(x)) -> f(c(f(x)))
      f(f(x)) -> f(d(f(x)))
    Qed